數(shù)值分析第五版答案

1、第一章第一章緒論緒論p192設(shè)的相對(duì)誤差為2%，求的相對(duì)誤差。xnx解：設(shè)，則函數(shù)的條件數(shù)為()nfxx?()||()pxfxCfx?又1()nfxnx???1||npxnxCnn?????又(())()rprxnCx?????且為2%()rex(())0.02nrxn???5計(jì)算球體積要使相對(duì)誤差限為1，問(wèn)度量半徑R時(shí)允許的相對(duì)誤差限是多少？解：球體體積為343VR??則何種函數(shù)的條件數(shù)為234343pRVRRCVR?????AA()

2、()3()rprrVCRR??????A又()1rV???故度量半徑R時(shí)允許的相對(duì)誤差限為1()10.333rR????7求方程的兩個(gè)根，使它至少具有4位有效數(shù)字（）。25610xx???78327.982?解：，25610xx???故方程的根應(yīng)為1228783x??故1287832827.98255.982x?????具有5位有效數(shù)字1x?2111287830.0178632827.98255.98228783x????????具有5

3、位有效數(shù)字2x解：由表格知，01234012340.40.50.60.70.8()0.916291()0.693147()0.510826()0.356675()0.223144xxxxxfxfxfxfxfx???????????????若采用線性插值法計(jì)算即，ln0.54(0.54)f則0.50.540.6??2112122111122()10(0.6)()10(0.5)()()()()()xxlxxxxxxlxxxxLxfxlxfx

4、lx??????????????6.93147(0.6)5.10826(0.5)xx????1(0.54)0.62021860.620219L?????若采用二次插值法計(jì)算時(shí)，ln0.541200102021101201220212001122()()()50(0.5)(0.6)()()()()()100(0.4)(0.6)()()()()()50(0.4)(0.5)()()()()()()()()()xxxxlxxxxxxxxxxxl

5、xxxxxxxxxxxlxxxxxxxLxfxlxfxlxfxlx????????????????????????????500.916291(0.5)(0.6)69.3147(0.4)(0.6)0.51082650(0.4)(0.5)xxxxxx????????????2(0.54)0.615319840.615320L?????4設(shè)xj為互異節(jié)點(diǎn)，求證：（1）0()nkkjjjxlxx???(01)kn??（2）0()()0nkjj